The probability of a reaction occurring. Why do chemical reactions occur? Variants of tasks for independent solution

(from the Greek kineticos driving) the science of the mechanisms of chemical reactions and the patterns of their occurrence over time. At 19 V. As a result of the development of the fundamentals of chemical thermodynamics, chemists learned to calculate the composition of an equilibrium mixture for reversible chemical reactions. In addition, based on simple calculations, it was possible, without conducting experiments, to draw a conclusion about the fundamental possibility or impossibility of a specific reaction occurring under given conditions. However, it is “possible in principle”reaction does not mean that it will go. For example, the reaction C + O 2 ® CO 2 from a thermodynamic point of view, it is very favorable, at least at temperatures below 1000° C (at higher temperatures the decomposition of CO molecules occurs 2 ), i.e. carbon and oxygen should (with almost 100% yield) turn into carbon dioxide. However, experience shows that a piece of coal can lie in the air for years, with free access to oxygen, without undergoing any changes. The same can be said about many other known reactions. For example, mixtures of hydrogen with chlorine or with oxygen can persist for a very long time without any signs of chemical reactions, although in both cases the reactions are thermodynamically favorable. This means that after reaching equilibrium in the stoichiometric mixture H 2 + Cl 2 Only hydrogen chloride should remain, and in the mixture 2H 2 + O 2 water only. Another example: acetylene gas is quite stable, although the reaction C 2 H 2 ® 2C + H 2 not only thermodynamically allowed, but also accompanied by a significant release of energy. Indeed, at high pressures, acetylene explodes, but under normal conditions it is quite stable.

Thermodynamically allowed reactions like those considered can only occur under certain conditions. For example, after ignition, coal or sulfur spontaneously combines with oxygen; hydrogen reacts easily with chlorine when the temperature rises or when exposed to ultraviolet light; a mixture of hydrogen and oxygen (explosive gas) explodes when ignited or when a catalyst is added. Why do all these reactions require special influences such as heating, irradiation, and the action of catalysts? Chemical thermodynamics does not answer this question; the concept of time is absent in it. At the same time, for practical purposes it is very important to know whether a given reaction will take place in a second, in a year, or over many millennia.

Experience shows that the speed of different reactions can differ greatly. Many reactions occur almost instantly in aqueous solutions. Thus, when an excess of acid is added to an alkaline solution of crimson-colored phenolphthalein, the solution instantly becomes discolored, which means that the neutralization reaction, as well as the reaction of converting the colored form of the indicator into a colorless one, proceed very quickly. The oxidation reaction of an aqueous solution of potassium iodide with atmospheric oxygen proceeds much more slowly: the yellow color of the reaction product iodine appears only after a long time. Corrosion processes of iron and especially copper alloys, as well as many other processes, occur slowly.

Predicting the rate of a chemical reaction, as well as elucidating the dependence of this rate on the reaction conditions is one of the important tasks of chemical kinetics, a science that studies the patterns of reactions over time. No less important is the second task facing chemical kinetics - the study of the mechanism of chemical reactions, that is, the detailed path of transformation of starting substances into reaction products.

Speed reaction. The easiest way to determine the rate is for a reaction occurring between gaseous or liquid reagents in a homogeneous (homogeneous) mixture in a vessel of constant volume. In this case, the reaction rate is defined as the change in the concentration of any of the substances participating in the reaction (it can be the starting substance or the reaction product) per unit time. This definition can be written as a derivative: v = d c/d t, Where v reaction speed; t time, c concentration. This speed is easy to determine if there is experimental data on the dependence of the concentration of the substance on time. Using this data, you can construct a graph called a kinetic curve. The rate of reaction at a given point on the kinetic curve is determined by the slope of the tangent at that point. Determining the slope of a tangent always involves some error. The initial reaction rate is most accurately determined, since at first the kinetic curve is usually close to a straight line; this makes it easier to draw a tangent at the starting point of the curve.

If time is measured in seconds, and concentration in moles per liter, then the reaction rate is measured in units of mol/(l

· With). Thus, the reaction rate does not depend on the volume of the reaction mixture: under the same conditions, it will be the same in a small test tube and in a large-scale reactor.

Value d

t is always positive, whereas the sign of d c depends on how the concentration changes over time: it decreases (for starting substances) or increases (for reaction products). To ensure that the reaction rate always remains a positive value, in the case of starting substances a minus sign is placed in front of the derivative: v = d c/d t . If the reaction occurs in the gas phase, pressure is often used instead of the concentration of substances in the rate equation. If the gas is close to ideal, then the pressure R is related to concentration with a simple equation: p = cRT. During a reaction, different substances can be consumed and formed at different rates, according to the coefficients in the stoichiometric equation ( cm. STOICHIOMETRY), therefore, when determining the rate of a specific reaction, these coefficients should be taken into account. For example, in the synthesis reaction of ammonia 3H 2 + N 2 ® 2NH 3 Hydrogen is consumed 3 times faster than nitrogen, and ammonia accumulates 2 times faster than nitrogen is consumed. Therefore, the rate equation for this reaction is written as follows: v = 1/3 d p(H2)/d t= d p(N 2)/d t= +1/2d p(NH 3)/d t . In general, if the reaction is stoichiometric, i.e. proceeds exactly in accordance with the written equation: aA + b B ® cC + dD, its speed is determined as v = (1/a)d[A]/d t= (1/b)d[B]/d t= (1/c)d[C]/d t= (1/d)d[D]/d t (square brackets are used to indicate the molar concentration of substances). Thus, the rates for each substance are strictly related to each other and, having determined experimentally the rate for any participant in the reaction, it is easy to calculate it for any other substance.

Most reactions used in industry are heterogeneous-catalytic. They occur at the interface between the solid catalyst and the gas or liquid phase. At the interface between two phases, reactions such as roasting of sulfides, dissolution of metals, oxides and carbonates in acids, and a number of other processes also occur. For such reactions, the rate also depends on the size of the interface, therefore the rate of a heterogeneous reaction is related not to a unit volume, but to a unit surface area. Measure the surface area on which the reaction occurs

It's not always easy.

If a reaction occurs in a closed volume, then its speed in most cases is maximum at the initial moment of time (when the concentration of the starting substances is maximum), and then, as the starting reagents are converted into products and, accordingly, their concentration decreases, the reaction rate decreases. There are also reactions in which the rate increases with time. For example, if a copper plate is immersed in a solution of pure nitric acid, the reaction rate will increase over time, which is easy to observe visually. The processes of dissolution of aluminum in alkali solutions, oxidation of many organic compounds with oxygen, and a number of other processes also accelerate over time. The reasons for this acceleration may be different. For example, this may be due to the removal of a protective oxide film from the metal surface, or to the gradual heating of the reaction mixture, or to the accumulation of substances that accelerate the reaction (such reactions are called autocatalytic).

In industry, reactions are usually carried out by continuously feeding starting materials into the reactor and removing products. Under such conditions, it is possible to achieve a constant rate of chemical reaction. Photochemical reactions also proceed at a constant rate, provided that the incident light is completely absorbed ( cm. PHOTOCHEMICAL REACTIONS).

Limiting stage of the reaction. If a reaction is carried out through sequential stages (not necessarily all of them are chemical) and one of these stages requires much more time than the others, that is, it proceeds much more slowly, then this stage is called limiting. It is this slowest stage that determines the speed of the entire process. Let us consider as an example the catalytic reaction of ammonia oxidation. There are two possible limiting cases here.

1. The flow of reactant molecules, ammonia and oxygen, to the surface of the catalyst (physical process) occurs much more slowly than the catalytic reaction itself on the surface. Then, to increase the rate of formation of the target product, nitrogen oxide, it is completely useless to increase the efficiency of the catalyst, but care must be taken to accelerate the access of the reagents to the surface.

2. The supply of reagents to the surface occurs much faster than the chemical reaction itself. This is where it makes sense to improve the catalyst, to select optimal conditions for the catalytic reaction, since the limiting stage in this case is the catalytic reaction on the surface.

Collision theory. Historically, the first theory on the basis of which the rates of chemical reactions could be calculated was the collision theory. Obviously, in order for molecules to react, they must first collide. It follows that the reaction should proceed faster, the more often the molecules of the starting substances collide with each other. Therefore, every factor that affects the frequency of collisions between molecules will also affect the rate of reaction. Some important laws concerning collisions between molecules were obtained on the basis of the molecular kinetic theory of gases.

In the gas phase, molecules move at high speeds (hundreds of meters per second) and very often collide with each other. The frequency of collisions is determined primarily by the number of particles per unit volume, that is, concentration (pressure). The frequency of collisions also depends on temperature (as it increases, molecules move faster) and on the size of molecules (large molecules collide with each other more often than small ones). However, concentration has a much stronger effect on collision frequency. At room temperature and atmospheric pressure, each medium-sized molecule experiences several billion collisions per second.

® C between two gaseous compounds A and B, assuming that a chemical reaction occurs whenever reactant molecules collide. Let there be a mixture of reagents A and B at equal concentrations in a liter flask at atmospheric pressure. There will be 6 in total in the flask· 10 23 /22.4 = 2.7 · 10 22 molecules, of which 1.35· 10 22 molecules of substance A and the same number of molecules of substance B. Let each molecule A experience 10 in 1 s 9 collisions with other molecules, of which half (5· 10 8 ) occurs in collisions with molecules B (collisions A + A do not lead to a reaction). Then in total 1.35 occur in the flask in 1 s· 10 22 · 5 · 10 8 ~ 7 · 10 30 collisions of molecules A and B. Obviously, if each of them led to a reaction, it would take place instantly. However, many reactions proceed quite slowly. From this we can conclude that only a tiny fraction of collisions between reactant molecules leads to interaction between them.

To create a theory that would allow one to calculate the reaction rate based on the molecular kinetic theory of gases, it was necessary to be able to calculate the total number of collisions of molecules and the proportion of “active” collisions leading to reactions. It was also necessary to explain why the rate of most chemical reactions increases greatly with increasing temperature the speed of molecules and the frequency of collisions between them increase slightly with temperature proportionally

, that is, only 1.3 times with an increase in temperature from 293 K (20° C) up to 373 K (100 ° C), while the reaction rate can increase thousands of times.

These problems were solved based on collision theory as follows. During collisions, molecules continuously exchange velocities and energies. Thus, as a result of a “successful” collision, a given molecule can noticeably increase its speed, while in an “unsuccessful” collision it can almost stop (a similar situation can be observed in the example of billiard balls). At normal atmospheric pressure, collisions, and therefore changes in speed, occur with each molecule billions of times per second. In this case, the velocities and energies of the molecules are largely averaged. If at a given moment in time we “recount” molecules with certain speeds in a given volume of gas, it turns out that a significant part of them have a speed close to the average. At the same time, many molecules have a speed less than the average, and some move at speeds greater than the average. As speed increases, the fraction of molecules having a given speed quickly decreases. According to collision theory, only those molecules that, when colliding, have a sufficiently high speed (and, therefore, a large supply of kinetic energy) react. This assumption was made in 1889 by a Swedish chemist Svante Arrhenius

. Activation energy. Arrhenius introduced into use by chemists the very important concept of activation energy ( Ea ) this is the minimum energy that a molecule (or a pair of reacting molecules) must have in order to enter into a chemical reaction. Activation energy is usually measured in joules and is referred not to one molecule (this is a very small value), but to a mole of a substance and is expressed in units of J/mol or kJ/mol. If the energy of the colliding molecules is less than the activation energy, then the reaction will not take place, but if it is equal to or greater, then the molecules will react.

Activation energies for different reactions are determined experimentally (from the dependence of the reaction rate on temperature). The activation energy can vary over a fairly wide range, from units to several hundred kJ/mol. For example, for the reaction 2NO

2 ® N 2 O 4 activation energy is close to zero for the 2H reaction 2 O 2 ® 2H 2 O + O 2 in aqueous solutions Ea = 73 kJ/mol, for thermal decomposition of ethane into ethylene and hydrogen Ea = 306 kJ/mol.

The activation energy of most chemical reactions significantly exceeds the average kinetic energy of molecules, which at room temperature is only about 4 kJ/mol and even at a temperature of 1000

° C does not exceed 16 kJ/mol. Thus, in order to react, molecules usually must have a speed much greater than average. For example, in case Ea = 200 kJ/mol colliding molecules of small molecular weight should have a speed of the order of 2.5 km/s (the activation energy is 25 times greater than the average energy of molecules at 20° WITH). And this is a general rule: for most chemical reactions, the activation energy significantly exceeds the average kinetic energy of the molecules.

The probability for a molecule to accumulate large energy as a result of a series of collisions is very small: such a process requires for it a colossal number of successive “successful” collisions, as a result of which the molecule only gains energy without losing it. Therefore, for many reactions, only a tiny fraction of molecules have sufficient energy to overcome the barrier. This share, in accordance with the Arrhenius theory, is determined by the formula:

a = e E a/ RT= 10 E a/2.3 RT~10 E a/19 T, where R = 8.31 J/(mol. TO). From the formula it follows that the proportion of molecules with energy Ea , as well as the proportion of active collisions a , very strongly depends on both the activation energy and temperature. For example, for a reaction with Ea = 200 kJ/mol at room temperature ( T~ 300 K) the fraction of active collisions is negligible: a = 10 200000/(19 , 300) ~ 10 35. And if every second 7 things happen in the vessel· 10 30 collisions of molecules A and B, it is clear that the reaction will not take place.

If you double the absolute temperature, i.e. heat the mixture to 600 K (327 ° C); At the same time, the proportion of active collisions will increase sharply:

a = 10 200000/(19 , 600) ~ 4·10 18 . Thus, a 2-fold increase in temperature increased the proportion of active collisions by 4 10 17 once. Now every second out of a total of approximately 7 10 30 collisions will lead to a reaction 7 10 30 4 10 18 ~ 3 10 13 . A reaction in which every second 3 10 13 molecules (out of about 10 22 ), although very slowly, it still goes. Finally, at a temperature of 1000 K (727 ° C) a ~ 3·10 11 (out of every 30 billion collisions of a given reactant molecule, one will result in a reaction). This is already a lot, since in 1 s 7 10 30 3 10 11 = 2 10 20 molecules, and such a reaction will take place in a few minutes (taking into account the decrease in the frequency of collisions with a decrease in the concentration of reagents).

Now it is clear why increasing the temperature can increase the rate of a reaction so much. The average speed (and energy) of molecules increases slightly with increasing temperature, but the proportion of “fast” (or “active”) molecules that have a sufficient speed of motion or sufficient vibrational energy for a reaction to occur increases sharply.

Calculation of the reaction rate, taking into account the total number of collisions and the fraction of active molecules (i.e., activation energy), often gives satisfactory agreement with experimental data. However, for many reactions the experimentally observed rate turns out to be less than that calculated by collision theory. This is explained by the fact that for a reaction to occur, the collision must be successful not only energetically, but also “geometrically,” that is, the molecules must be oriented in a certain way relative to each other at the moment of the collision. Thus, when calculating reaction rates using collision theory, in addition to the energy factor, the steric (spatial) factor for a given reaction is also taken into account.

Arrhenius equation. The dependence of the reaction rate on temperature is usually described by the Arrhenius equation, which in its simplest form can be written as v = v 0 a = v 0 e E a/ RT , Where v 0 the speed that the reaction would have at zero activation energy (in fact, this is the frequency of collisions per unit volume). Because the v 0 weakly depends on temperature, everything is determined by the second factor exponential: with increasing temperature this factor increases rapidly, and the faster the higher the activation energy E A. This dependence of the reaction rate on temperature is called the Arrhenius equation; it is one of the most important in chemical kinetics. To approximate the effect of temperature on the reaction rate, the so-called “van’t Hoff rule” is sometimes used ( cm. Van't Hoff's Rule).

If a reaction obeys the Arrhenius equation, the logarithm of its rate (measured, for example, at the initial moment) should linearly depend on the absolute temperature, that is, the plot of ln

v from 1/ T must be straightforward. The slope of this line is equal to the activation energy of the reaction. Using such a graph, you can predict what the reaction rate will be at a given temperature or at what temperature the reaction will proceed at a given speed. Several practical examples of using the Arrhenius equation.

1. The packaging of a frozen product says that it can be stored on a refrigerator shelf (5° C) for 24 hours, in a freezer marked with one star (6° C) for a week, two stars (12° C) for a month. , and in a freezer with a *** symbol (which means the temperature in it is 18 ° C) 3 months. Assuming that the rate of product spoilage is inversely proportional to the guaranteed shelf life

t xp, in ln coordinates t хр , 1/ T we obtain, in accordance with the Arrhenius equation, a straight line. From it you can calculate the activation energy of biochemical reactions leading to spoilage of a given product (about 115 kJ/mol). From the same graph you can find out to what temperature the product must be cooled so that it can be stored, for example, 3 years; it turns out to be 29° C.

2. Mountaineers know that in the mountains it is difficult to boil an egg, or in general any food that requires more or less long boiling. Qualitatively, the reason for this is clear: with a decrease in atmospheric pressure, the boiling point of water decreases. Using the Arrhenius equation, you can calculate how long it will take, for example, to hard boil an egg in Mexico City, located at an altitude of 2265 m, where the normal pressure is 580 mm Hg, and water at such a reduced pressure boils at 93 ° C The activation energy for the protein “folding” (denaturation) reaction was measured and turned out to be very large compared to many other chemical reactions - about 400 kJ/mol (it may differ slightly for different proteins). In this case, lowering the temperature from 100 to 93 ° C (that is, from 373 to 366 K) will slow down the reaction by 10

(400000/19)(1/366 1/373) = 11.8 times. This is why residents of the highlands prefer frying food to cooking: the temperature of a frying pan, unlike the temperature of a pan of boiling water, does not depend on atmospheric pressure.

3. In a pressure cooker, food is cooked at increased pressure and, therefore, at an increased boiling point of water. It is known that in a regular saucepan, beef is cooked for 23 hours, and apple compote for 1015 minutes. Considering that both processes have similar activation energies (about 120 kJ/mol), we can use the Arrhenius equation to calculate that in a pressure cooker at 118°C the meat will cook for 2530 minutes, and the compote for only 2 minutes.

The Arrhenius equation is very important for the chemical industry. When an exothermic reaction occurs, the released thermal energy heats not only the environment, but also the reactants themselves. this may result in an undesirable rapid acceleration of the reaction. Calculating the change in reaction rate and heat release rate with increasing temperature allows us to avoid a thermal explosion ( cm. EXPLOSIVE SUBSTANCES).

Dependence of the reaction rate on the concentration of reagents. The rate of most reactions gradually decreases over time. This result is in good agreement with the collision theory: as the reaction proceeds, the concentrations of the starting substances fall, and the frequency of collisions between them decreases; Accordingly, the frequency of collisions of active molecules decreases. This leads to a decrease in the reaction rate. This is the essence of one of the basic laws of chemical kinetics: the rate of a chemical reaction is proportional to the concentration of reacting molecules. Mathematically, this can be written as the formula v = k[A][B], where k a constant called the reaction rate constant. The equation given is called the chemical reaction rate equation or kinetic equation. The rate constant for this reaction does not depend on the concentration of the reactants and on time, but it depends on temperature in accordance with the Arrhenius equation: k = k 0 e E a/ RT . The simplest speed equation v = k [A][B] is always true in the case when molecules (or other particles, for example, ions) A, colliding with molecules B, can directly transform into reaction products. Such reactions, occurring in one step (as chemists say, in one stage), are called elementary reactions. There are few such reactions. Most reactions (even seemingly simple ones like H 2 + I 2 ® 2HI) are not elementary, therefore, based on the stoichiometric equation of such a reaction, its kinetic equation cannot be written.

The kinetic equation can be obtained in two ways: experimentally by measuring the dependence of the reaction rate on the concentration of each reagent separately, and theoretically if the detailed reaction mechanism is known. Most often (but not always) the kinetic equation has the form

v = k[A] x[B] y , Where x and y are called reaction orders for reactants A and B. These orders, in the general case, can be integer and fractional, positive and even negative. For example, the kinetic equation for the reaction of thermal decomposition of acetaldehyde CH 3 CHO ® CH 4 + CO has the form v = k 1,5 , i.e. the reaction is one and a half order. Sometimes a random coincidence of stoichiometric coefficients and reaction orders is possible. Thus, the experiment shows that the reaction H 2 + I 2 ® 2HI is first order in both hydrogen and iodine, that is, its kinetic equation has the form v = k(This is why this reaction was considered elementary for many decades, until its more complex mechanism was proven in 1967).

If the kinetic equation is known, i.e. It is known how the reaction rate depends on the concentrations of the reactants at each moment of time, and the rate constant is known, then it is possible to calculate the time dependence of the concentrations of the reactants and reaction products, i.e. theoretically obtain all kinetic curves. For such calculations, methods of higher mathematics or computer calculations are used, and they do not present any fundamental difficulties.

On the other hand, the experimentally obtained kinetic equation helps to judge the reaction mechanism, i.e. about a set of simple (elementary) reactions. Elucidation of reaction mechanisms is the most important task of chemical kinetics. This is a very difficult task, since the mechanism of even a seemingly simple reaction can include many elementary stages.

The use of kinetic methods to determine the reaction mechanism can be illustrated using the example of alkaline hydrolysis of alkyl halides to form alcohols: RX +

OH ® ROH + X . It was experimentally discovered that for R = CH 3, C 2 H 5 etc. and X = Cl, the reaction rate is directly proportional to the concentrations of the reactants, i.e. has the first order in the halide RX and the first in the alkali, and the kinetic equation has the form v = k 1 . In the case of tertiary alkyl iodides (R = (CH 3) 3 C, X = I) order in RX first, and in alkali zero: v = k 2 . In intermediate cases, for example, for isopropyl bromide (R = (CH 3) 2 CH, X = Br), the reaction is described by a more complex kinetic equation: v = k 1 + k 2 . Based on these kinetic data, the following conclusion was made about the mechanisms of such reactions.

In the first case, the reaction occurs in one step, through direct collision of alcohol molecules with OH ions

(the so-called SN mechanism 2 ). In the second case, the reaction occurs in two stages. First stage slow dissociation of the alkyl iodide into two ions: R I ® R + + I . Second very fast reaction between ions: R+ + OH ® ROH. The rate of the total reaction depends only on the slow (limiting) stage, so it does not depend on the alkali concentration; hence zero order in alkali (SN mechanism 1 ). In the case of secondary alkyl bromides, both mechanisms occur simultaneously, so the kinetic equation is more complex.

Ilya Leenson

LITERATURE History of the doctrine of the chemical process. M., Nauka, 1981
Leenson I.A. Chemical reactions. M., AST Astrel, 2002

The first law of thermodynamics allows us to calculate the thermal effects of various processes, but does not provide information about the direction of the process.

For processes occurring in nature, two driving forces are known:

1. The desire of the system to move to a state with the least amount of energy;

2. The desire of the system to achieve the most probable state, which is characterized by the maximum number of independent particles.

The first factor is characterized by a change in enthalpy. The case under consideration must be accompanied by the release of heat, therefore, DH< 0.

The second factor is determined by temperature and change entropy.

Entropy (S)- thermodynamic function of the state of the system, which reflects the probability of the implementation of a particular state of the system in the process of heat exchange.

Like energy, entropy is not an experimentally determined quantity. In a reversible process occurring under isothermal conditions, the change in entropy can be calculated using the formula:

This means that during an irreversible process, entropy increases due to the conversion of part of the work into heat.

Thus, in reversible processes the system performs the maximum possible work. In an irreversible process, the system always does less work.

The transition of lost work into heat is a feature of heat as a macroscopically disordered form of energy transfer. This gives rise to the interpretation of entropy as a measure of disorder in a system:

With increasing disorder in a system, entropy increases and, conversely, with ordering of the system, entropy decreases.

Thus, in the process of water evaporation, entropy increases, and in the process of water crystallization, it decreases. In decomposition reactions, entropy increases, in connection reactions it decreases.

The physical meaning of entropy was established by statistical thermodynamics. According to Boltzmann's equation:

The direction of the spontaneous occurrence of the process depends on the ratio of the quantities on the left and right sides of the last expression.

If the process takes place under isobaric-isothermal conditions, then the overall driving force of the process is called Gibbs free energy or isobaric-isothermal potential (DG):

(15)

The DG value allows you to determine the direction of the spontaneous occurrence of the process:

If DG< 0, то процесс самопроизвольно протекает в прямом направлении;

If DG > 0, then the process spontaneously proceeds in the opposite direction;

If DG=0, then the state is equilibrium.

In living organisms, which are open systems, the main source of energy for many biological reactions - from protein biosynthesis and ion transport to muscle contraction and electrical activity of nerve cells - is ATP (adenosine-5¢-triphosphate).

Energy is released during ATP hydrolysis:

ATP + H 2 O ⇄ ADP + H 3 PO 4

where ADP is adenosine-5¢-diphosphate.

DG 0 of this reaction is -30 kJ, therefore the process proceeds spontaneously in the forward direction.

Analysis of the relationship between the enthalpy and entropy factors in the equation for calculating the isobaric-isothermal potential allows us to draw the following conclusions:

1. At low temperatures, the enthalpy factor predominates, and exothermic processes occur spontaneously;

2. At high temperatures, the entropy factor predominates, and processes accompanied by an increase in entropy occur spontaneously.

Based on the material presented, we can formulate II law of thermodynamics:

Under isobaric-isothermal conditions in an isolated system, those processes that are accompanied by an increase in entropy occur spontaneously.

Indeed, in an isolated system, heat exchange is impossible, therefore, DH = 0 and DG » -T×DS. This shows that if the DS value is positive, then the DG value is negative and, therefore, the process spontaneously proceeds in the forward direction.

Another formulation of the Second Law of Thermodynamics:

An uncompensated transfer of heat from less heated bodies to more heated ones is impossible.

In chemical processes, changes in entropy and Gibbs energy are determined in accordance with Hess's law:

,	(16)
.	(17)

Reactions for which DG< 0 называют exergonic.

Reactions for which DG > 0 are called endergonic.

The DG value of a chemical reaction can also be determined from the relationship:

DG = DH - T×DS.

In table Figure 1 shows the possibility (or impossibility) of a spontaneous reaction for various combinations of signs DH and DS.

Problem solving standards

1. Some reaction occurs with a decrease in entropy. Determine under what conditions the spontaneous occurrence of this reaction is possible.

The condition for the spontaneous occurrence of a reaction is a decrease in the Gibbs free energy, i.e. DG< 0. Изменение DG можно рассчитать по формуле:

Since entropy decreases during the reaction (DS< 0), то энтропийный фактор препятствует самопроизвольному протеканию данной реакции. Таким образом, самопроизвольное протекание данной реакции может обеспечить только энтальпийный фактор. Для этого необходимо выполнение следующих условий:

1) DH< 0 (реакция экзотермическая);

2) (the process must take place at low temperatures).

2. The endothermic decomposition reaction occurs spontaneously. Estimate the change in enthalpy, entropy and Gibbs free energy.

1) Since the reaction is endothermic, DH > 0.

2) In decomposition reactions, entropy increases, therefore DS > 0.

3) The spontaneous occurrence of the reaction indicates that DG< 0.

3. Calculate the standard enthalpy of chemosynthesis occurring in the bacteria Thiobacillus denitrificans:

6KNO 3(solid) + 5S (solid) + 2CaCO 3(solid) = 3K 2 SO 4(solid) + 2CaSO 4(solid) + 2CO 2(gas) + 3N 2(gas)

according to the values of standard enthalpies of formation of substances:

Let us write down the expression of the first corollary from Hess’s law, taking into account the fact that the standard enthalpies of formation of sulfur and nitrogen are equal to zero:

= (3× K 2 SO 4 + 2× CaSO 4 + 2× CO 2) -

- (6× KNO 3 + 2× CaCO 3).

Let us substitute the values of the standard enthalpies of formation of substances:

3×(-1438) + 2×(-1432) + 2×(-393.5) - (6×(-493) + 2×(-1207)).

2593 kJ.

Because< 0, то реакция экзотермическая.

4. Calculate the standard enthalpy of the reaction:

2C 2 H 5 OH (liquid) = C 2 H 5 OC 2 H 5 (liquid) + H 2 O (liquid)

according to the values of standard enthalpies of combustion of substances:

C 2 H 5 OH = -1368 kJ/mol;

C 2 H 5 OC 2 H 5 = -2727 kJ/mol.

Let us write the expression for the second corollary from Hess’s law, taking into account the fact that the standard enthalpy of combustion of water (higher oxide) is zero:

2× C 2 H 5 OH - C 2 H 5 OC 2 H 5 .

Let us substitute the values of the standard enthalpies of combustion of the substances participating in the reaction:

2×(-1368) - (-2727).

Corollaries from Hess's law make it possible to calculate not only the standard enthalpies of reactions, but also the values of the standard enthalpies of formation and combustion of substances using indirect data.

5. Determine the standard enthalpy of formation of carbon monoxide (II) using the following data:

From equation (1) it can be seen that the standard change in enthalpy of this reaction corresponds to the standard enthalpy of formation of CO 2.

Let us write down the expression of the first corollary from Hess’s law for reaction (2):

CO = CO 2 - .

Let's substitute the values and get:

CO = -293.5 - (-283) = -110.5 kJ/mol.

This problem can be solved in another way.

Subtracting the second from the first equation, we get:

6. Calculate the standard entropy of the reaction:

CH 4 (gas) + Cl 2 (gas) = CH 3 Cl (gas) + HCl (gas),

according to the values of standard entropies of substances:

We calculate the standard entropy of the reaction using the formula:

= (CH 3 Cl + HCl) - (CH 4 + Cl 2).

234 + 187 - (186 + 223) = 12 J/(mol×K).

7. Calculate the standard Gibbs energy of the reaction:

C 2 H 5 OH (liquid) + H 2 O 2 (liquid) = CH 3 COH (gas) + 2H 2 O (liquid)

according to the following data:

Determine whether spontaneous occurrence of this reaction is possible under standard conditions.

We calculate the standard Gibbs energy of the reaction using the formula:

= (CH 3 COH + 2× H 2 O) - (C 2 H 5 OH + H 2 O 2).

Substituting the table values, we get:

129 + 2×(-237) - ((-175) + (-121) = -307 kJ/mol.

Because< 0, то самопроизвольное протекание данной реакции возможно.

C 6 H 12 O 6 (solid) + 6O 2 (gas) = 6CO 2 (gas) + 6H 2 O (liquid).

according to known data:

We calculate the values of the standard enthalpy and entropy of the reaction using the first corollary of Hess’s law:

6 CO 2 + 6 H 2 O - C 6 H 12 O 6 - 6 O 2 =

6×(-393.5) + 6×(-286) - (-1274.5) - 6×0 = -2803 kJ;

6 CO 2 + 6 H 2 O - C 6 H 12 O 6 - 6 O 2 =

6x214 + 6x70 - 212 - 6x205 = 262 J/K = 0.262 kJ/K.

We find the standard Gibbs energy of the reaction from the relation:

T× = -2803 kJ - 298.15 K×0.262 kJ/K =

9. Calculate the standard Gibbs energy of the hydration reaction of serum albumin at 25 0 C, for which DH 0 = -6.08 kJ/mol, DS 0 = -5.85 J/(mol×K). Assess the contribution of the enthalpy and entropy factors.

We calculate the standard Gibbs energy of the reaction using the formula:

DG 0 = DH 0 - T×DS 0.

Substituting the values, we get:

DG 0 = -6.08 kJ/mol - 298 K×(-5.85×10 - 3) kJ/(mol×K) =

4.34 kJ/mol.

In this case, the entropy factor prevents the reaction from occurring, and the enthalpy factor favors it. Spontaneous reaction is possible provided that , i.e., at low temperatures.

10. Determine the temperature at which the trypsin denaturation reaction spontaneously occurs, if = 283 kJ/mol, = 288 J/(mol×K).

The temperature at which both processes are equally probable can be found from the relation:

In this case, the enthalpy factor prevents the reaction from occurring, and the entropy factor favors it. Spontaneous reaction is possible provided that:

Thus, the condition for the spontaneous occurrence of the process is T > 983 K.

Questions for self-control

1. What is a thermodynamic system? What types of thermodynamic systems do you know?

2. List the thermodynamic parameters known to you. Which ones are measured? Which ones are unmeasurable?

3. What is a thermodynamic process? What are the names of processes that occur when one of the parameters is constant?

4. What processes are called exothermic? Which ones are endothermic?

5. What processes are called reversible? Which ones are irreversible?

6. What is meant by the term “system state”? What are the system states?

7. What systems does classical thermodynamics study? Formulate the first and second postulates of thermodynamics.

8. What variables are called state functions? List the state functions that you know.

9. What is internal energy? Is it possible to measure internal energy?

10. What is enthalpy? What is its dimension?

11. What is entropy? What is its dimension?

12. What is Gibbs free energy? How can it be calculated? What can you determine using this function?

13. What reactions are called exergonic? Which ones are endergonic?

14. Formulate the first law of thermodynamics. What is the equivalence between heat and work?

15. Formulate Hess’s law and consequences from it. What is the standard enthalpy of formation (combustion) of a substance?

16. Formulate the second law of thermodynamics. Under what conditions does a process occur spontaneously in an isolated system?

Variants of tasks for independent solution

Option #1

4NH 3 (gas) + 5O 2 (gas) = 4NO (gas) + 6H 2 O (gas),

Determine what type (exo- or endothermic) this reaction belongs to.

C 2 H 6 (gas) + H 2 (gas) = 2CH 4 (gas),

3. Calculate the standard Gibbs energy of the hydration reaction of b-lactoglobulin at 25 0 C, for which DH 0 = -6.75 kJ, DS 0 = -9.74 J/K. Assess the contribution of the enthalpy and entropy factors.

Option No. 2

1. Calculate the standard enthalpy of the reaction:

2NO 2 (gas) + O 3 (gas) = O 2 (gas) + N 2 O 5 (gas),

using the values of standard enthalpies of formation of substances:

Determine what type (exo- or endothermic) this reaction belongs to.

2. Calculate the standard enthalpy of the reaction:

using the values of standard enthalpies of combustion of substances:

3. Calculate the standard Gibbs energy of the reaction of thermal denaturation of chymotrypsinogen at 50 0 C, for which DH 0 = 417 kJ, DS 0 = 1.32 J/K. Assess the contribution of the enthalpy and entropy factors.

Option No. 3

1. Calculate the standard enthalpy of the reaction of benzene hydrogenation to cyclohexane in two ways, i.e., using the values of the standard enthalpies of formation and combustion of substances:

Cu (solid) + ZnO (solid) = CuO (solid) + Zn (solid)

3. When 12.7 g of copper (II) oxide is reduced by coal (to form CO), 8.24 kJ of heat is absorbed. Determine the standard enthalpy of formation of CuO if CO = -111 kJ/mol.

Option No. 4

1. Calculate the standard enthalpy of chemosynthesis occurring in the autotrophic bacteria Baglatoa and Thiothpix, by stages and in total:

2H 2 S (gas) + O 2 (gas) = 2H 2 O (liquid) + 2S (solid);

2S (solid) + 3O 2 (gas) + 2H 2 O (liquid) = 2H 2 SO 4 (liquid),

2. Calculate the standard enthalpy of the reaction:

C 6 H 12 O 6 (solid) = 2C 2 H 5 OH (liquid) + 2CO 2 (gas),

using the values of standard enthalpies of combustion of substances:

4HCl (gas) + O 2 (gas) = 2Cl 2 (gas) + 2H 2 O (liquid)

according to known data:

Option #5

1. Calculate the standard enthalpy of the reaction:

2CH 3 Cl (gas) + 3O 2 (gas) = 2CO 2 (gas) + 2H 2 O (liquid) + 2HCl (gas),

using the values of standard enthalpies of formation of substances:

Determine what type (exo- or endothermic) this reaction belongs to.

2. Calculate the standard enthalpy of the reaction:

C 6 H 6 (liquid) + 3H 2 (gas) = C 6 H 12 (liquid),

using the values of standard enthalpies of combustion of substances:

3. Calculate the standard Gibbs energy of the trypsin denaturation reaction at 50 0 C, for which DH 0 = 283 kJ, DS 0 = 288 J/K). Assess the possibility of the process occurring in the forward direction.

Option #6

1. Calculate the standard enthalpy of chemosynthesis occurring in the autotrophic bacteria Thiobacillus Thioparus:

5Na 2 S 2 O 3 × 5H 2 O (solid) + 7O 2 (gas) = 5Na 2 SO 4 (solid) + 3H 2 SO 4 (liquid) + 2S (solid) + 22H 2 O (liquid .) ,

Determine what type (exo- or endothermic) this reaction belongs to.

2. Calculate the standard enthalpy of the reaction:

C 6 H 5 NO 2 (liquid) + 3H 2 (gas) = C 6 H 5 NH 2 (liquid) + 2H 2 O (liquid),

using the values of standard enthalpies of combustion of substances:

3. Assess the role of enthalpy and entropy factors for the reaction:

H 2 O 2 (liquid) + O 3 (gas) = 2O 2 (gas) + H 2 O (liquid)

according to known data:

Determine the temperature at which the reaction will proceed spontaneously.

Option No. 7

1. Calculate the standard enthalpy of formation of CH 3 OH using the following data:

CH 3 OH (liquid) + 1.5O 2 (gas) = CO 2 (gas) + 2H 2 O (liquid) DH 0 = -726.5 kJ;

C (graphite) + O 2 (gas) = CO 2 (gas) DH 0 = -393.5 kJ;

H 2 (gas) + 0.5O 2 (gas) = H 2 O (liquid) DH 0 = -286 kJ.

2. Assess the possibility of a spontaneous reaction:

8Al (sol.) + 3Fe 3 O 4 (sol.) = 9Fe (sol.) + Al 2 O 3 (sol.)

under standard conditions, if:

3. Calculate the value of DH 0 for possible reactions of glucose conversion:

1) C 6 H 12 O 6 (cr.) = 2C 2 H 5 OH (liquid) + 2CO 2 (gas);

2) C 6 H 12 O 6 (red) + 6O 2 (gas) = 6CO 2 (gas) + 6H 2 O (liquid).

according to known data:

Which of these reactions releases the most energy?

Option No. 8

1. Calculate the standard enthalpy of formation of MgCO 3 using the following data:

MgO (solid) + CO 2 (gas) = MgCO 3 (solid) +118 kJ;

C 2 H 6 (gas) + H 2 (gas) = 2CH 4 (gas)

according to known data:

3. Which of the following oxides: CaO, FeO, CuO, PbO, FeO, Cr 2 O 3 can be reduced by aluminum to free metal at 298 K:

Option No. 9

1. Calculate the standard enthalpy of formation of Ca 3 (PO 4) 2 using the following data:

3CaO (tv.) + P 2 O 5 (tv.) = Ca 3 (PO 4) 2 (tv.) DH 0 = -739 kJ;

P 4 (tv.) + 5O 2 (gas) = 2P 2 O 5 (tv.) DH 0 = -2984 kJ;

Ca (solid) + 0.5O 2 (gas) = CaO (solid) DH 0 = -636 kJ.

2. Assess the possibility of a spontaneous reaction:

Fe 2 O 3 (solid) + 3CO (gas) = 2Fe (solid) + 3CO 2 (gas)

under standard conditions, if:

3. Determine which of the following oxides: CuO, PbO 2, ZnO, CaO, Al 2 O 3 can be reduced by hydrogen to free metal at 298 K, if it is known:

Option No. 10

1. Calculate the standard enthalpy of formation of ethanol using the following data:

DH 0 combustion C 2 H 5 OH = -1368 kJ/mol;

C (graphite) + O 2 (gas) = CO 2 (gas) +393.5 kJ;

H 2 (gas) + O 2 (gas) = H 2 O (liquid) +286 kJ.

2. Calculate the standard entropy of the reaction:

C 2 H 2 (gas) + 2H 2 (gas) = C 2 H 6 (gas),

according to known data:

3. Calculate the amount of energy that will be released in the body of a person who ate 2 pieces of sugar, 5 g each, assuming that the main way of sucrose metabolism is reduced to its oxidation:

C 12 H 22 O 11 (solid) + 12O 2 (gas) = 12CO 2 (gas) + 11H 2 O (liquid) = -5651 kJ.

Option No. 11

1. Calculate the standard enthalpy of formation of C2H4 using the following data:

C 2 H 4 (gas) + 3O 2 (gas) = 2CO 2 (gas) + 2H 2 O (liquid) +1323 kJ;

C (graphite) + O 2 (gas) = CO 2 (gas) +393.5 kJ;

H 2 (gas) + 0.5O 2 (gas) = H 2 O (liquid) +286 kJ.

2. Without making calculations, set the sign DS 0 of the following processes:

1) 2NH 3 (gas) = N 2 (gas) + 3H 2 (gas);

2) CO 2 (cr.) = CO 2 (gas);

3) 2NO (gas) + O 2 (gas) = 2NO 2 (gas).

3. Determine by what reaction equation the decomposition of hydrogen peroxide will proceed under standard conditions:

1) H 2 O 2 (gas) = H 2 (gas) + O 2 (gas);

2) H 2 O 2 (gas) = H 2 O (liquid) + 0.5 O 2 (gas),

Option No. 12

1. Calculate the standard enthalpy of formation of ZnSO 4 using the following data:

2ZnS + 3O 2 = 2ZnO + SO 2 DH 0 = -890 kJ;

2SO 2 + O 2 = 2SO 3 DH 0 = -196 kJ;

H 2 O (solid) = H 2 O (liquid),

H 2 O (liquid) = H 2 O (gas),

H 2 O (solid) = H 2 O (gas).

according to known data:

3. Calculate the amount of energy that will be released during the combustion of 10 g of benzene, using the following data:

Option No. 14

1. Calculate the standard enthalpy of formation of PCl 5 from the following data:

P 4 (solid) + 6Cl 2 (gas) = 4PCl 3 (gas) DH 0 = -1224 kJ;

PCl 3 (gas) + Cl 2 (gas) = PCl 5 (gas) DH 0 = -93 kJ.

2. Calculate the standard change in the Gibbs energy of formation of carbon disulfide CS 2 using the following data:

CS 2 (liquid) + 3O 2 (gas) = CO 2 (gas) + 2SO 2 (gas) DG 0 = -930 kJ;

CO 2 = -394 kJ/mol; SO 2 = -300 kJ/mol.

3. Assess the role of enthalpy and entropy factors for the reaction:

CaCO 3 (solid) = CaO (solid) + CO 2 (gas)

according to known data:

Determine the temperature at which the reaction will proceed spontaneously.

Option No. 15

1. Calculate the thermal effect of the reaction of formation of crystalline hydrate CuSO 4 × 5H 2 O, proceeding according to the equation:

CuSO 4 (solid) + 5H 2 O (liquid) = CuSO 4 × 5H 2 O (solid) ,

Enthalpy- this is a quantity that characterizes the energy reserve in a substance.

Enthalpy is also called heat content. The greater the energy reserve, the greater the enthalpy of the substance.

The thermal effect of the reaction (at constant pressure) is equal to the change in enthalpy (ΔH):

For an exothermic reaction Q > 0, ΔH< 0, поскольку относительно реагентов энергия теряется в окружающую среду. И наоборот, для эндотермической реакции Q < 0, ΔН >0 - energy is acquired from the environment.

By analogy with the standard heat of formation Q o6p, there is also the concept of a standard enthalpy of formation, which is denoted ΔH arr. Its values are given in reference tables.

The thermochemical equation of the same reaction can be written in different ways:

The human body is a unique “chemical reactor” in which many different chemical reactions take place. Their main difference from the processes occurring in a test tube, flask, or industrial plant is that in the body all reactions occur under “mild” conditions (atmospheric pressure, low temperature), and few harmful by-products are formed.

The process of oxidation of organic compounds with oxygen is the main source of energy in the human body, and its main end products are carbon dioxide CO 2 and water H 2 O.

For example:

This released energy is a large quantity, and if food were oxidized quickly and completely in the body, then just a few pieces of sugar eaten would cause the body to overheat. But biochemical processes, the total thermal effect of which, according to Hess’s law, does not depend on the mechanism and is a constant value, proceed stepwise, as if extended in time. Therefore, the body does not “burn out”, but economically spends this energy on vital processes. But does this always happen?

Every person should have at least an approximate idea of how much energy enters his body with food and how much is consumed during the day.

One of the principles of rational nutrition is this: the amount of energy supplied from food should not exceed energy consumption (or be less) by more than 5%, otherwise metabolism is disrupted and a person becomes fat or loses weight.

The energy equivalent of food is its calorie content, expressed in kilocalories per 100 g of product (often indicated on the packaging, can also be found in special reference books and cookbooks). And energy consumption in the body depends on age, gender, and intensity of work.

The most beneficial food is one that is low in calories, but contains all the components in the food (proteins, fats, carbohydrates, minerals, vitamins, microelements).

The energy value of food and the calorific value of fuel are associated with exothermic reactions of their oxidation. The driving force of such reactions is the “striving” of the system towards a state with the lowest internal energy.

Exothermic reactions begin spontaneously, or require only a small “push”—an initial supply of energy.

What then is the driving force behind endothermic reactions, during which thermal energy comes from the environment and is stored in the reaction products, turning into their internal energy? This is due to the tendency of any system to the most probable state, which is characterized by maximum disorder, it is called entropy. For example, the molecules that make up the air do not fall to the Earth, although the minimum potential energy of each molecule corresponds to its lowest position, since the desire for the most probable state causes the molecules to be randomly distributed in space.

Imagine that you poured different nuts into a glass. It is almost impossible to achieve separation and order by shaking them, since in this case the system will tend to the most probable state in which disorder in the system increases, so the nuts will always be mixed. Moreover, the more particles we have, the greater the likelihood of disorder.

The greatest order in chemical systems is in an ideal crystal at absolute zero temperature. They say that entropy in this case is zero. With increasing temperature in the crystal, random vibrations of atoms (molecules, ions) begin to intensify. Entropy increases. This occurs especially sharply at the moment of melting during the transition from a solid to a liquid, and even more so at the moment of evaporation during the transition from liquid to gas.

The entropy of gases significantly exceeds the entropy of liquids and, especially, solids. If you spill a little gasoline in an enclosed space, such as a garage, you will soon smell it throughout the entire room. Evaporation (endothermic process) and diffusion occur, a random distribution of gasoline vapor throughout the entire volume. Gasoline vapor has greater entropy compared to liquid.

The process of boiling water from an energy point of view is also an endothermic process, but it is beneficial from the point of view of an increase in entropy when the liquid transforms into vapor. At a temperature of 100 °C, the entropy factor “outweighs” the energy factor - water begins to boil - water vapor has greater entropy compared to liquid water.

As you analyze the data in Table 12, notice how low the entropy value is for a diamond that has a very regular structure. Substances formed by more complex particles have higher entropy values. For example, the entropy of ethane is greater than the entropy of methane.

Table 12
Some values of standard molar entropy

Spontaneous endothermic reactions are precisely those reactions in which a fairly strong increase in entropy is observed, for example due to the formation of gaseous products from liquid or solid substances or due to an increase in the number of particles.

For example:

CaCO 3 → CaO + CO 2 - Q,

2NH 3 → N 2 + ZN 2 - Q.

Let us formulate conclusions.

The direction of a chemical reaction is determined by two factors: the desire to reduce internal energy with the release of energy and the desire for maximum disorder, i.e., to increase entropy.
An endothermic reaction can be made to proceed if it is accompanied by an increase in entropy.
Entropy increases with increasing temperature and especially strongly during phase transitions: solid - liquid, solid - gaseous.
The higher the temperature at which the reaction is carried out, the greater the importance of the entropy factor compared to the energy factor.

There are experimental and theoretical methods for determining the entropies of various chemical compounds. Using these methods, it is possible to quantify the changes in entropy during a particular reaction, in a similar way to the thermal effect of a reaction. As a result, it becomes possible to predict the direction of a chemical reaction (Table 13).

Table 13
The possibility of chemical reactions occurring depending on changes in energy and entropy

To answer the question about the possibility of a reaction, a special quantity was introduced - the Gibbs energy (G), which allows us to take into account both the change in enthalpy and the change in entropy:

ΔG = ΔН - TΔS,

where T is the absolute temperature.

Only those processes occur spontaneously in which the Gibbs energy decreases, i.e., the value ΔG< 0. Процессы, при которых ΔG >0 are basically impossible. If ΔG = 0, i.e. ΔН = TΔS, then chemical equilibrium has been established in the system (see § 14).

Let's return to case No. 2 (see Table 13).

All life on our planet - from viruses and bacteria to humans - consists of highly organized matter, which is more ordered compared to the surrounding world. For example, protein. Remember its structures: primary, secondary, tertiary. You are already well acquainted with the “substance of heredity” (DNA), the molecules of which consist of structural units arranged in a strictly defined sequence. This means that protein or DNA synthesis is accompanied by a huge decrease in entropy.

In addition, the initial building material for the growth of plants and animals is formed in the plants themselves from water H 2 O and carbon dioxide CO 2 during the process of photosynthesis:

6H 2 O + 6CO 2(G) → C6H 12 O 6 + 6O 2(G).

In this reaction, entropy decreases and a reaction occurs with the absorption of light energy. This means the process is endothermic! Thus, the reactions to which we owe our lives turn out to be thermodynamically forbidden. But they are coming! And this uses the energy of light quanta in the visible region of the spectrum, which is much greater than thermal energy (infrared quanta). In nature, endothermic reactions with a decrease in entropy, as you can see, occur under certain conditions. Chemists cannot yet create such conditions artificially.

Questions and tasks for § 12

Basic concepts of chemical thermodynamics

Chemical processes can occur with a change in the chemical composition of a substance (chemical reactions) or without changing it (phase transitions). A set of substances that interact and are isolated from the surrounding space (mentally) is called system. For example: a hydrogen atom (a system of a nucleus and an electron), an aqueous solution of various salts, etc.

Depending on the nature of the interaction of the system with the environment, there are: open or open (there is an exchange of heat, energy and matter with the environment), closed or closed (there is an exchange of heat and energy with the environment, but there is no exchange of matter) and isolated(lack of mass and heat transfer between the system and the environment) (Fig. 1).

Rice. 1. Examples of closed (a), open (b) and isolated systems (c).

The state of the system is determined by the totality of its properties and is characterized by thermodynamic parameters: temperature, pressure and volume (T, p, V). Any change in one or more parameters of a system is called a thermodynamic process. Thus, an increase in temperature leads to a change in the internal energy of the system (U).

DEFINITION

Internal energy– the total supply of molecules, atoms, electrons and nuclei that make up the system, consisting of the kinetic energy of these particles and the energy of interaction between them.

The absolute value of U cannot be calculated or measured. It is possible to determine the change in internal energy (ΔU) as a result of a process. ΔU of any system when transitioning from one state to another does not depend on the transition path, but is determined by the initial and final positions of the system. This means that the internal energy of the system is a function of state.

ΔU = U 2 – U 1,

Where 1 and 2 are symbols of the initial and final state of the system.

The first law of thermodynamics: the heat Q imparted to the system is spent on increasing internal energy and performing work (A) against external forces:

It should be noted that A and Q are not state functions, i.e. do not depend on the path of the process.

In thermodynamics, quantities are often introduced that are identical to the sum of several thermodynamic parameters. This replacement greatly simplifies the calculations. Thus, the state function equal to U + pV is called enthalpy (H):

At constant pressure (isobaric process) and in the absence of other work except expansion work, heat is equal to the change in enthalpy:

Q p = ΔU + pΔV = ΔH

If the process occurs at a constant volume (isochoric) and in the absence of other work, the released or absorbed heat corresponds to the change in internal energy:

Basics of Thermochemistry

The branch of chemical thermodynamics that studies the heats of chemical reactions and their dependence on various physical and chemical parameters is called thermochemistry. In thermochemistry, thermochemical equations of reactions are used, in which the state of aggregation of a substance is indicated, and the thermal effect of the reaction is considered as one of the products. For example:

2H 2(g) + O 2(g) = H 2 O (g) + 242 kJ,

Which means that when 1 mole of water is formed in a gaseous state, 242 kJ of heat is released. In this case, the change in enthalpy ΔH = − 242 kJ.

The opposite signs of Q and ΔH indicate that in the first case this is a characteristic of processes in the environment, and in the second - in the system. For an exothermic process Q > 0, ΔH< 0, а при эндотермическом – наоборот.

Thermal effects can not only be measured, but also calculated using Hess's law: the thermal effect of a chemical reaction occurring at constant p and V does not depend on the number of intermediate stages, but is determined only by the initial and final state of the system.

Consequences of Hess's law

There are 5 consequences from Hess's law:

1) The thermal effect of the formation of 1 mole of a complex substance from simple substances, under standard conditions, is called the heat of formation of this substance - ΔH 0 f. So, for example, ΔH 0 f (CO 2) from C (s) and O 2 (g) will be equal to −393.51 kJ.

2) Standard heats of formation of simple substances are equal to zero.

3) The standard thermal effect of a chemical reaction (ΔH 0) is equal to the difference between the sum of the heats of formation of the reaction products (taking into account stoichiometric coefficients) and the sum of the heats of formation of the starting substances (including stoichiometric coefficients):

ΔH 0 = Σ ΔH 0 f (products) − Σ ΔH 0 f (reagents)

For example, for a reaction:

2H 2 S (g) + 3O 2 (g) = 2SO 2 (g) + 2H 2 O (aq)

ΔH 0 = Σ (2 × ΔH 0 f (SO 2) + 2 × ΔH 0 f (H 2 O)) – Σ (2 ΔH 0 f (H 2 S) +0)

4) The thermal effect of a chemical reaction is equal to the difference between the sum of the heats of combustion of the starting substances and the sum of the heats of combustion of the reaction products, taking into account stoichiometric coefficients

5) All algebraic operations can be performed with thermochemical equations, for example:

A= B + C + 400 kJ

B + D = A − 200 kJ

Adding these equations we get

A + B + D = B + C + A + 200 kJ

D = C + 200 kJ

ΔH 0 = − 200 kJ

Entropy. Direction of chemical processes. Gibbs energy

DEFINITION

Entropy (S)– a property of a system, the change in which during a reversible process is numerically equal to the ratio of heat to the temperature of the process:

For example, when water evaporates under boiling conditions (T = 373 K, p = 1 atm), the change in entropy is equal to ΔS = ΔH exp /373 = 44000/373 = 118 kJ/(mol × K).

Based on the standard entropy of substances (S 0), it is possible to calculate the change in entropy of various processes:

Δ r S 0 = Σ n i S 0 − Σ n j S 0 ,

where i are reaction products, j are starting materials.

The entropy of simple substances is not zero.

By calculating Δ r S 0 and Δ r H 0 we can conclude that the reaction is reversible. So, if Δ r S 0 and Δ r H 0 are greater than zero or Δ r S 0 and Δ r H 0 are less than zero, then the reaction is reversible.

There is a function that connects the change in enthalpy and entropy and answers the question about the spontaneity of the reaction - the Gibbs energy (G).

ΔG = ΔH − T × ΔS

Δ r G 0 = Δ r H 0 − T × Δ r S 0

The direction of a chemical reaction is judged by the value Δ r G 0 . If Δ r G 0<0, то реакция идет в прямом направлении, а если Δ r G 0 >0 – in reverse. Of the 2 reactions, the one with the lower Δ r G 0 value will proceed with the highest probability.

Table 1. Conditions for the spontaneous occurrence of chemical reactions

Examples of problem solving

EXAMPLE 1

Exercise

Calculate the change in Gibbs energy (Δ G o 298) for the process:

Na 2 O(s) + H 2 O(l) → 2NaOH(s)

Is it possible for a spontaneous reaction to occur under standard conditions and 298K?

Required reference data: Δ G o f (NaOH,t) = –381.1 kJ/mol, Δ G o f (Na 2 O) = –378 kJ/mol, Δ G o f (H 2 O, l) = –237 kJ/mol.

Solution

Under standard conditions and T=298K Δ G o 298 can be calculated as the difference in the total Gibbs energy (Δ G o f) formation of reaction products and the total Gibbs energy of formation of starting substances:

Δ G o 298 = 2Δ G o f (NaOH,t) – [Δ G o f (Na 2 O, t) + Δ G o f (H 2 O, l)]

Δ G o 298 = 2(–381.1) –[–378 + (–237)] = –147.2 kJ.

Δ value G o 298 is negative, so spontaneous reaction is possible.

Answer

Δ G o 298 = –147.2 kJ, spontaneous reaction is possible.

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